RIGID BODY

A rigid body is a body with a definite and unchanged shape and size i.e. a body is said to be rigid if the distance between any two particles of the body remains invariant.

Motion of A Rigid Body

Translational: If a body moves such that its orientation does not change with respect to time then body is said to move in translational motion.

ROTATION: If a body is rotating about the fix axis of rotation then its motion is known as pure rotational motion.

MIX – MOTION: If a body moves such that its motion neither be pure rotational nor be pure translational then its motion is known as mix motion.

Angular Displacement

Consider a rigid body undergoing rotation about an axis, perpendicular to the plane of the paper and passing through O. Suppose that A and B are any two particles of the rigid body at the position 1 while A' and B' are their subsequent locations when the body is at the position 2.

Since the body undergoes rotation,

OA = OA' and OB = OB'

Further AB = A'B', since the body is rigid.

$\Rightarrow$ $\Delta$OAB $\cong$ $\Delta$OA'B'(congruent)

i.e. $\angle$ AOB = $\angle$A'OB'

Adding $\angle$ AOB' to both sides of the above equation, we get

$\angle$ BOB' = $\angle$AOA' =$\theta$(say)

This implies that in a given interval of time the angular displacements of all particles of the rigid body undergoing rotation are identical.

Therefore, a single variable, viz. angular displacement ($\theta$) can be used to describe the rotational motion of the rigid body. Angular displacement is not a vector quantity

ANGULAR VELOCITY($\omega$)

The rate of change of angular displacement with respect to time is known as angular velocity.

Average angular velocity: The rate of change of angular displacement with respect to time is known as angular velocity.

To define average angular velocity it is necessary to specify the interval in which we are talking about average.

$\omega$= $\frac{{\theta }_2-{\theta }_1}{t_2-t_1}$

Instantaneous angular velocity: Instantaneous velocity means angular velocity at a particular instant. It is mathematically defined as ,

$\omega =\underset{t_2\to t_1}{\lim }\frac{{\theta }_2-{\theta }_1}{t_2-t_1}=\frac{d\theta }{dt}]att=t_1$

ANGULAR ACCELERATIONS ($\alpha$)

Angular accelerations ($\alpha$): The rate of change of angular velocity with respect to time is known as angular acceleration.

Average angular acceleration: It is necessary to specify the time interval to in which we are talking about average.

In a given time interval t₁ to t₂ the average angular acceleration is defined as , $\vec{\alpha }=\frac{{\vec{\omega }}_2-{\vec{\omega }}_1}{t_2-t_1}$

Instantaneous angular acceleration: Instantaneous angular acceleration means angular acceleration at a particular dot instant at t = t₁ mathematically it is defined as, $\vec{\alpha }={\underset{t_1\to t_2}{\mathrm{limit}}\frac{{\vec{\omega }}_2-{\vec{\omega }}_1}{t_2-t_1}=\frac{d\omega }{dt}]}_{t=t_2}$

Direction of angular acceleration: If magnitude of$\omega$increasing then direction of $\alpha$ will be same as direction of$\omega$and vice–versa.

EQUATION OF ANGULAR MOTION

$\omega$(t) = $\omega$_o + $\alpha$t

$\theta$(t) = $\theta$_o + $\omega$_ot + $\frac{1}{2}$$\alpha$t²$\omega$${}_{\left(t\right)}^2$ =${\omega }_0^2+2\alpha \theta$

Here $\omega$_o = magnitude of the initial angular velocity

$\omega$(t) = magnitude of the angular velocity after time t.

$\theta$_o=Initial angular position.

$\theta$(t) = Angular position after time t.

Illustration 1: A disc starts rotating with constant angular acceleration of $\pi$radian/s² about a fixed axis perpendicular to its plane and passing through its centre.Find

(a) The angular velocity of the disc after 4 sec.

(b) The angular displacement of the disc after 4 sec and

(c) Number of turns accomplished by the disc in 4 sec.

Solution: Here $\alpha$= $\pi$rad/sec²

$\omega$₀ = 0 t = 4 sec

(a) $\omega$_{(4 sec)} = 0 + ($\pi$rad/sec²) x 4 sec= 4$\pi$rad/sec.

(b) $\theta$_{(4 sec)} = 0 + $\frac{1}{2}$($\pi$rad/sec²) x (16 sec²) = 8 $\pi$radian.

(c) Let the number of turns be n

$\Rightarrow$ n 2 $\pi$rad = 8 $\pi$rad $\Rightarrow$ n = 4

RELATION BETWEEN LINEAR AND ANGULAR VARIABLES

Consider a particle A of a rigid body undergoing rotation about a fixed axis-$\hat{n}$, the particle A describing an arc ABA' of a circle with its centre O on the axis of rotation. Taking the origin at O, the position vector of A,

$\vec{r}$ $\equiv$ $\overrightarrow{\mathrm{OA}}$

OA = OA' = constant(radius of the circle)

$\angle$A'OA =$\theta$(t) (say)

The arc length, ABA', S = r$\theta$

The tangential velocity, v_A = $\frac{dS}{dt}=\frac{d\theta }{dt}r$ = r$\omega$

The direction of the angular velocity vector $\vec{\omega }$ be taken along the axis of rotation:

$\vec{\omega }$= $\omega$$\hat{n}$, $\hat{n}$ being the unit vector along the axis of rotation.Then, $\vec{v}$, instantaneous velocity of A with respect to the axis of rotation, can be written as ,$\vec{v}=\vec{\omega }\times \vec{r}$The acceleration of the point A with respect to the axis of rotation is$\Rightarrow$$\vec{a}$= $\frac{d\vec{v}}{dt}=\frac{d\vec{\omega }}{dt}\times \vec{r}+\vec{\omega }\times \frac{d\vec{r}}{dt}$ $\Rightarrow$$\vec{a}=\vec{\alpha }\times \vec{r}+\vec{\omega }\times \vec{v}$ If $\vec{\omega }$ is constant, then$\vec{\alpha }$ = 0 and $\vec{a}=\vec{\omega }\times \vec{v}$ = $\vec{\omega }$x$\left(\vec{\omega }\times \vec{r}\right)$= - ${\omega }^2\vec{r}$

TORQUE

Torque of a force about the axis of rotation: The turning effect of a force about the axis of rotation is called moment of force or torque due to the force and is measured as the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the axis of rotation. It is denoted by the Greek letter $\tau$(tou). Thus,torque = force x perpendicular distance from the axis of rotationIts unit is N m in SI. Its dimensional formula is [M L² T^–2], same as that of work. It may be pointed out that no doubt, N m is equivalent to joule (the unit of work); but joule is not used as the unit of torque. Consider a force $\overrightarrow{F}$ acting on a particle P. Choose an origin O and let $\overrightarrow{r}$ be the position vector of the particle experiencing the force. We define the torque of the force F about O as , $\overrightarrow{P}=\overrightarrow{r}\times \overrightarrow{F}$ … (1) This is a vector quantity having its direction perpendicular to $\overrightarrow{r}$ and $\overrightarrow{F}$ according to the rule of cross product. Now consider a rigid body rotating about a given axis of rotation AB (fig b). Let F be a force acting on the particle O of the body. F may not be in the plane ABP. Take the origin O somewhere on the axis of rotation. The torque of F about O is $\overrightarrow{P}=\overrightarrow{r}\times \overrightarrow{F}$. Its component along OA is called the torque of $\overrightarrow{F}$ about OA. To calculate it, we should find the vector $\overrightarrow{r}\times \overrightarrow{F}$ and then find out the angle$\theta$it makes with OA. The torque about OA is then $\left|\overrightarrow{r}\times \overrightarrow{F}\right|\cos \theta$. The torque of a force about a line. This can be shown as given below. Let O₁ be any point on the line AB (Figure 10.4). The torque of F about O₁ is ${\vec{O}}_{1}P\times \vec{F}=({\vec{O}}_{1}O+\vec{O}P)\times \vec{F}={\vec{O}}_{1}O\times \vec{F}+\vec{O}P\times \vec{F}$.As ${\vec{O}}_{1}P\times \vec{F}\perp {\vec{O}}_{1}O,$ this term will have no component along AB.Thus, the component of ${\vec{O}}_{1}P\times \vec{F}$ is equal to that of ${\vec{O}}_{1}P\times \vec{F}$.
Illustration 2: A particle of mass m is dropped at point A, find the torque about O. Solution: $\vec{\tau }=$$\vec{r}\times \vec{F}$$\Rightarrow$ $\vec{\tau }=$ r F sin$\theta$ $\hat{n}$= (r sin$\theta$) F$\hat{n}$

$\Rightarrow$ $\tau$= b mg

The direction of torque is directed inward the paper or in other words, rotation about O is clockwise.

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KINETIC ENERGY OF A RIGID BODY

Let a rigid body is purely rotating about an axis AB with angular velocity$\omega$consider a general particle $\Delta$m₂ which is at a distance of r₂ from axis of rotation.

V₂ = r₂.$\omega$

So energy associated with this $\Delta$m₂ is $\Delta$K · E₂

$\Delta$K · E₂ = $\frac{1}{2}\Delta m_2{v}_2^2=\frac{1}{2}\Delta m_2{r}_2^2{\omega }^2$

K·E_total = $\sum \Delta k\cdot E_2=\sum \frac{1}{2}\Delta m_2\cdot r_2^2{\omega }^2={\omega }^2\sum \frac{1}{2}\Delta m_2{r}_2^2$

K·E_total = _{$\frac{1}{2}I{\omega }^2$.}

Where I is called as moment of inertia of body about an given axis of rotation. In this case I is about AB.

Moment of inertia is also called as rotational mass of object.

Illustration 3: Four point masses each of value m are placed at points A, B, C and D at distances $\frac{a}{2},a,\frac{3a}{2}$ and 2a from the free and of a mass-less rod. (i) Find the M.I. of the system about an axis perpendicular to the point B. (ii) Will the mass on the left side contribute a negative term?

Solution: Suppose that the four masses each of mass m are placed at the points A, B, C and D of a mass-less rod (fig.)

(i) M.I. of the rod about the axis through the point B,

I = m (BA)² + m (BB)² + m (BC)² + m (BD)²

= m (a/2)² + m (0)² + m (a/2)² + m (a)²

= $\frac{1}{2}m{a}^2+0+\frac{1}{4}m{a}^2+m{a}^2=\frac{3}{4}m{a}^2$

(ii) The distance occurs with power 2 in the formula for moment of inertia.

Therefore, if the distance of the mass on left of the axis is taken as negative, its moment of inertia will still contribute a positive term. Hence, the mass m on the left side of the axis will not contribute a negative term to the moment of inertia of the system.